[soupastupid]

1. calculate the equilibrium constant for the reaction
H_{2 (g)}  +  I_{2 (g)}   ⇋   2 HI _(g)
at 80.0 °C. The vapor pressure of solid iodine is 0.0216 bar at that temperature. If 42.7 g of solid iodine are placed in a 10.0 L vessel at 80.0 °C, what is the minimum amount of hydrogen gas that must be introduced in order to remove all the solid iodine? (_fH°298 (I2(g)) = 62.44 kJ mol^-1, S°298 (I_2(g)) = 260.69 J mol^-1K^-1)


2. Relevant equations
ΔrG = ΔfG(products) - ΔfG(reactants)
ln K = -ΔrG / RT
K= [HI]^2 / [H_2]*[I_2]

ln K = -ΔH/RT + ΔS/T

3. The attempt at a solution

ΔfG for HI(g) = 1.70, for H2(g) = 0 and for I2(g) = 19.33 KJ mol-1
ΔrG = 2(1.7) - 0 - 19.33
ΔrG = -15.93 KJ mol-1
ln K = -ΔrG / RT
ln K = -(-15.93E3) / (8/314)(353) = 5.42
K = 225.9

K= [HI]^2 / [H_2]*[I_2]

lnK = -62.44jk/mol/ (8.314 J/kmol)*(80oC) + 260.69 J/molK / 8.314 J/kmol
K= 2.22*10^4 for converting solid iodine to gas

now what?
i tried using mass balance with 42.7g Iodine
i got 0.168 mol I2 and H2
and 0.336 mol of HI

but i dont think its right because i dont know how to use the 10.0 L or 0.0216 bar
do i use PV=nRT??
0.0216 bar * 10.0 L = 0.168 mol I2 * 0.08206 LatmK-1mol-1 * 80oC??? no unknowns???
and they dont even equal

[aznbaby949]

you need to convert the bar to atm.

and so what is your equilibrium constant for the reaction overall?

[soupastupid]

0.0216 bar = 0.0213 atm

i multiply the two equilibrium constants together to get

5.44 * 10^6

now what?

[Kemi]

You don't need the equilibrium constant for the sublimation of iodine (it would be the vapor pressure divided by the unit as the activity of a pure solid is 1).

It might be useful to think through the chemistry before attempting any calculations. At first you have solid and gaseous iodine at equilibrium. What happens when you add hydrogen gas? What happens eventually when you keep adding it?